The vapour pressure of a solvent decreased by `10 mm` of `Hg` when a non-volatile solute was added to the solvent. The mole fraction of solute is `0.2`, what would be the mole fraction of solvent if the decrease in vapour pressure is `20 mm` of `Hg`.

Options

A. `0.2`
B. `0.6`
C. `0.4`
D. `0.8`

Answer

2

Explanation

According to second Raoult's law, the relative lowering of vapour pressure of volatile solvent is equal to the mole fraction of the nonvolatile solute:
`(P_(1)^(0)-P_(1))/P_(1)^(0)= chi_(2)`
or `(DeltaP)/P_(1)^(0)= chi_(2)`
For the first solution
`(10 mm Hg)/P_(1)^(@)=0.2`
Thus, `P_(1)^(0)=50 mm Hg`
For the second solution
`chi_(2)=(20 mm Hg)/(50 mm Hg)=0.4`
Since `chi_(1)+chi_(2)=0`.
`chi_(1)=1- chi_(2)=1- 0.4= 0.6`